A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. When released

Question

A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. When released, the block moves on a horizontal tabletop for 1.0m before coming to rest. The force constant k is 100 N/m. What is the coefficient of kinetic friction between the block and tabletop

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Acacia 5 years 2021-08-04T13:22:31+00:00 1 Answers 122 views 0

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  1. taiduc
    0
    2021-08-04T13:24:09+00:00
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    Answer:

    So coefficient of kinetic friction will be equal to 0.4081

    Explanation:

    We have given mass of the block m = 0.5 kg

    The spring is compressed by length x = 0.2 m

    Spring constant of the sprig k = 100 N/m

    Blocks moves a horizontal distance of s = 1 m

    Work done in stretching the spring is equal to W=\frac{1}{2}kx^2=\frac{1}{2}\times 100\times 0.2^2=2J

    This energy will be equal to kinetic energy of the block

    And this kinetic energy must be equal to work done by the frictional force

    So \mu mg\times s=2

    \mu\times 0.5\times 9.8\times 1=2

    \mu =0.4081

    So coefficient of kinetic friction will be equal to 0.4081

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